N个人中每组只有一个领导者的组数

给定N人，任务是计算形成大小为? N其中，在每个组中，该组的第一个元素是该组的领导者。
笔记：

• 具有不同领导者的相同人员的组被视为不同的组。例如：组 {1, 2, 3} 和 {2, 1, 3} 被视为不同的组，因为它们分别具有不同的领导者 1 和 2。
• 具有相同领导者和相同人员的组被视为同一组。例如：组 {1, 3, 2} 和 {1, 2, 3} 被视为同一组，因为它们具有相同的领导者和相同的人。
• 答案可以很大，取模到(1e9+7)。

例子：

Input: N = 3 
Output: 12 
Explanation: 
Total Groups with leaders are: 
Groups with Leader 1: 
1. {1} 
2. {1, 2} 
3. {1, 3} 
4. {1, 2, 3} 
Groups with Leader 2: 
5. {2} 
6. {2, 1} 
7. {2, 3} 
8. {2, 1, 3} 
Groups with Leader 3: 
9. {3} 
10. {3, 1} 
11. {3, 2} 
12. {3, 1, 2}
Input: N = 5 
Output: 80

方法：这个问题可以使用二项式系数和模幂的概念来解决。以下是对该问题陈述的观察：

• 在N个人中选择一个领导者的方法数是C(N, 1) 。
• 对于每个领导者，我们可以选择一个大小为K的组，其中0 ≤ K ≤ N-1来进行分组的可能数量。
• 因此，总路数由N与从剩余(N – 1) 个元素中选择K 个元素的总和的乘积给出：

  Total Ways = $\binom{N}{1} * \left ( \binom{N-1}{0} + \binom{N-1}{1} + \binom{N-1}{2} + ... + \binom{N-1}{N-1} \right )$

通过使用二项式定理，二项式系数的和可以写成：

$\binom{N}{1} * \left ( \binom{N-1}{0} + \binom{N-1}{1} + \binom{N-1}{2} + ... + \binom{N-1}{N-1} \right ) = 2^{N-1}$

因此选择只有一个领导者的组的方法数是

$N * 2^{N - 1}$

下面是上述方法的实现：

C++

// C++ program for the above approach
  
#include 
using namespace std;
  
long long mod = 1000000007;
  
// Function to find 2^x using
// modular exponentiation
int exponentMod(int A, int B)
{
    // Base cases
    if (A == 0)
        return 0;
    if (B == 0)
        return 1;
  
    // If B is even
    long long y;
    if (B % 2 == 0) {
        y = exponentMod(A, B / 2);
        y = (y * y) % mod;
    }
  
    // If B is odd
    else {
        y = A % mod;
        y = (y * exponentMod(A, B - 1)
             % mod)
            % mod;
    }
  
    return (int)((y + mod) % mod);
}
  
// Function to count the number of
// ways to form the group having
// one leader
void countWays(int N)
{
  
    // Find 2^(N-1) using modular
    // exponentiation
    long long select = exponentMod(2,
                                   N - 1);
  
    // Count total ways
    long long ways
        = ((N % mod)
           * (select % mod));
  
    ways %= mod;
  
    // Print the total ways
    cout << ways;
}
  
// Driver Code
int main()
{
  
    // Given N number of peoples
    int N = 5;
  
    // Function Call
    countWays(N);
}

Java

// Java program for the above approach
import java.util.*;
class GFG{
  
static long mod = 1000000007;
  
// Function to find 2^x using
// modular exponentiation
static int exponentMod(int A, int B)
{
    // Base cases
    if (A == 0)
        return 0;
    if (B == 0)
        return 1;
  
    // If B is even
    long y;
    if (B % 2 == 0) 
    {
        y = exponentMod(A, B / 2);
        y = (y * y) % mod;
    }
  
    // If B is odd
    else 
    {
        y = A % mod;
        y = (y * exponentMod(A, B - 1) % 
                                  mod) % mod;
    }
  
    return (int)((y + mod) % mod);
}
  
// Function to count the number of
// ways to form the group having
// one leader
static void countWays(int N)
{
  
    // Find 2^(N-1) using modular
    // exponentiation
    long select = exponentMod(2, N - 1);
  
    // Count total ways
    long ways = ((N % mod) * (select % mod));
  
    ways %= mod;
  
    // Print the total ways
    System.out.print(ways);
}
  
// Driver Code
public static void main(String[] args)
{
  
    // Given N number of peoples
    int N = 5;
  
    // Function Call
    countWays(N);
}
}
  
// This code is contributed by sapnasingh4991

Python3

# Python3 program for the above approach
mod = 1000000007
  
# Function to find 2^x using
# modular exponentiation
def exponentMod(A, B):
      
    # Base cases
    if (A == 0):
        return 0;
    if (B == 0):
        return 1;
  
    # If B is even
    y = 0;
      
    if (B % 2 == 0):
        y = exponentMod(A, B // 2);
        y = (y * y) % mod;
  
    # If B is odd
    else:
        y = A % mod;
        y = (y * exponentMod(A, B - 1) %
                                  mod) % mod;
                               
    return ((y + mod) % mod);
  
# Function to count the number of
# ways to form the group having
# one leader
def countWays(N):
      
    # Find 2^(N-1) using modular
    # exponentiation
    select = exponentMod(2, N - 1);
  
    # Count total ways
    ways = ((N % mod) * (select % mod));
  
    ways %= mod;
  
    # Print the total ways
    print(ways)
      
# Driver code        
if __name__=='__main__':
      
    # Given N number of people
    N = 5;
  
    # Function call
    countWays(N);
  
# This code is contributed by rutvik_56

C#

// C# program for the above approach
using System;
class GFG{
   
static long mod = 1000000007;
   
// Function to find 2^x using
// modular exponentiation
static int exponentMod(int A, int B)
{
    // Base cases
    if (A == 0)
        return 0;
    if (B == 0)
        return 1;
   
    // If B is even
    long y;
    if (B % 2 == 0) 
    {
        y = exponentMod(A, B / 2);
        y = (y * y) % mod;
    }
   
    // If B is odd
    else
    {
        y = A % mod;
        y = (y * exponentMod(A, B - 1) % 
                                  mod) % mod;
    }
   
    return (int)((y + mod) % mod);
}
   
// Function to count the number of
// ways to form the group having
// one leader
static void countWays(int N)
{
   
    // Find 2^(N-1) using modular
    // exponentiation
    long select = exponentMod(2, N - 1);
   
    // Count total ways
    long ways = ((N % mod) * (select % mod));
   
    ways %= mod;
   
    // Print the total ways
    Console.Write(ways);
}
   
// Driver Code
public static void Main(String[] args)
{
   
    // Given N number of peoples
    int N = 5;
   
    // Function Call
    countWays(N);
}
}
  
// This code is contributed by sapnasingh4991

Javascript

  
// Javascript program for the above approach
  
let mod = 1000000007;
  
// Function to find 2^x using
// modular exponentiation
function exponentMod(A, B)
{
    // Base cases
    if (A == 0)
        return 0;
    if (B == 0)
        return 1;
  
    // If B is even
    let y;
    if (B % 2 == 0) {
        y = exponentMod(A, B / 2);
        y = (y * y) % mod;
    }
  
    // If B is odd
    else {
        y = A % mod;
        y = (y * exponentMod(A, B - 1)
            % mod)
            % mod;
    }
  
    return ((y + mod) % mod);
}
  
// Function to count the number of
// ways to form the group having
// one leader
function countWays(N)
{
  
    // Find 2^(N-1) using modular
    // exponentiation
    let select = exponentMod(2,
                                N - 1);
  
    // Count total ways
    let ways
        = ((N % mod)
        * (select % mod));
  
    ways %= mod;
  
    // Print the total ways
    document.write(ways);
}
  
// Driver Code
  
    // Given N number of peoples
    let N = 5;
  
    // Function Call
    countWays(N);
  
// This code is contributed by Mayank Tyagi
  

输出：

80

时间复杂度： O(log N)
辅助空间： O(1)

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